The sum of coefficients of all the integral powers of $x$ in the expansion of $(1 + 2 \sqrt{x})^{40}$ is

3^{40} + 1

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3^{40} - 1

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\frac{1}{2}

(3^{40} - 1)

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\frac{1}{2}

(3^{40} + 1)

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Solution

The correct option is D $\frac{1}{2}$ $(3^{40} + 1)$
$(1 + 2 \sqrt{x})^{40} = 1 +^{40} C_{1} (2 \sqrt{x})^{1} +^{40} C_{2} (2 \sqrt{x})^{2} +^{40} C_{3} (2 \sqrt{x})^{3} + . . . +^{40} C_{40} (2 \sqrt{x})^{40}$
Hence, all the odd terms will have integral powers of $x$ .
Substituting $x = 1$ , we get
$3^{40} = a_{0} + a_{1} + a_{2} + a_{3} . . . + a_{40}$ ...(i)
Now consider,
$(1 - 2 \sqrt{x})^{40} = 1 -^{40} C_{1} (2 \sqrt{x})^{1} +^{40} C_{2} (2 \sqrt{x})^{2} -^{40} C_{3} (2 \sqrt{x})^{3} . . . +^{40} C_{40} (2 \sqrt{x})^{40}$
Taking $x$ as $1$ , we get
$1 = a_{0} - a_{1} + a_{2} - a_{3} . . . + a_{40}$ ...(ii)
Adding i and, ii we get
$3^{40} + 1 = 2 [a_{0} + a_{2} + a_{4} + a_{6} . . . + a_{40}]$
$\frac{3^{40} + 1}{2} = a_{0} + a_{2} + a_{4} + a_{6} . . . + a_{40}$
Hence the sum of the coefficients, of integral powers of $x$ will be
$\frac{3^{40} + 1}{2}$
Hence, option D is correct

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